A) \[\frac{B}{4}\]
B) \[\frac{B}{2}\]
C) 2B
D) 4B
Correct Answer: B
Solution :
Magnetic field at the centre of current carrying loop \[B=\frac{{{\mu }_{0}}i}{2r}\] ?. (i) where r is the radius of loop and i the current, Given, \[{{r}_{2}}=2r\] \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{1}{r}\times \frac{2r}{1}=\frac{2}{1}\] \[\Rightarrow \] \[{{B}_{2}}=\frac{{{B}_{1}}}{2}=\frac{B}{2}\] Hence, magnetic field is halved.You need to login to perform this action.
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